A mass of 2.0, kg is placed on a flat pan att | vedclass

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(A) The spring has a natural length $l$. When mass $m$ is placed on it,the spring compresses by $x_0$ to reach the equilibrium position $O'$.At equili

Vedclass · Accepted Answer

$10\, cm$

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(A) The spring has a natural length $l$. When mass $m$ is placed on it,the spring compresses by $x_0$ to reach the equilibrium position $O'$.At equilibrium,the spring force balances the weight of the mass: $k x_0 = m g$.$x_0 = \frac{m g}{k} = \frac{2.0 	imes 10}{200} = 0.10\, m = 10\, cm$.When the mass oscillates with amplitude $A$,the maximum upward acceleration of the mass is $a_{max} = A \omega^2$,where $\omega^2 = \frac{k}{m}$.The mass will detach from the pan when the downward acceleration of the mass exceeds the acceleration due to gravity $g$ at the highest point of its motion,or equivalently,when the normal force becomes zero.The condition for the mass to lose contact with the pan is $A \omega^2 \ge g$.Substituting $\omega^2 = \frac{k}{m}$,we get $A (\frac{k}{m}) \ge g$.$A \ge \frac{m g}{k} = x_0$.Therefore,the minimum amplitude required for the mass to detach is $A = 10\, cm$.

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